Clustering and classification
This chapter focuses on performing and interpreting clustering and classification on the Boston data set.
Load the Boston data set
The data are for the housing values in suburbs of Boston. The data are available from the MASS package and the variable descriptions can be found here.
library(MASS)
data("Boston") # load the data
str(Boston) # A data frame## 'data.frame': 506 obs. of 14 variables:
## $ crim : num 0.00632 0.02731 0.02729 0.03237 0.06905 ...
## $ zn : num 18 0 0 0 0 0 12.5 12.5 12.5 12.5 ...
## $ indus : num 2.31 7.07 7.07 2.18 2.18 2.18 7.87 7.87 7.87 7.87 ...
## $ chas : int 0 0 0 0 0 0 0 0 0 0 ...
## $ nox : num 0.538 0.469 0.469 0.458 0.458 0.458 0.524 0.524 0.524 0.524 ...
## $ rm : num 6.58 6.42 7.18 7 7.15 ...
## $ age : num 65.2 78.9 61.1 45.8 54.2 58.7 66.6 96.1 100 85.9 ...
## $ dis : num 4.09 4.97 4.97 6.06 6.06 ...
## $ rad : int 1 2 2 3 3 3 5 5 5 5 ...
## $ tax : num 296 242 242 222 222 222 311 311 311 311 ...
## $ ptratio: num 15.3 17.8 17.8 18.7 18.7 18.7 15.2 15.2 15.2 15.2 ...
## $ black : num 397 397 393 395 397 ...
## $ lstat : num 4.98 9.14 4.03 2.94 5.33 ...
## $ medv : num 24 21.6 34.7 33.4 36.2 28.7 22.9 27.1 16.5 18.9 ...dim(Boston) ## [1] 506 14Comment: The data frame has 506 rows and 14 columns. All variables are numeric, with the variable chas: Charles River dummy variable (= 1 if tract bounds river; 0 otherwise).
A graphical overview of the data and summary of the variables
Matrix plot of the variables
# plot matrix of the variables
pairs(Boston,
col = "blue", # Change color
pch = 18, # Change shape of points
main = "Matrix plot of the variables") # Add a main title
The upper correlation matrix
library(corrplot) # access the corrplot library
# visualize the upper correlation matrix
corrplot(cor_matrix, method="circle", type = "upper")
Interpretations:
There seems to be a positive correlation between per capita crime rate by town (crim) and the index of accessibility to radial highways (rad) and also full-value property-tax rate per $10,000 (tax).
A slightly positive correlation between per capita crime rate by town (crim) with proportion of non-retail business acres per town (indus), nitrogen oxides concentration (parts per 10 million) (nox), and lower status of the population (percent) (lstat).
A positive correlation between the proportion of residential land zoned for lots over 25,000 square feet (zn) and the weighted mean of distances to five Boston employment centres (dis).
A positive correlation between the proportion of non-retail business acres per town (indus) with nitrogen oxides concentration (parts per 10 million) (nox), proportion of owner-occupied units built prior to 1940 (age), index of accessibility to radial highways (rad), full-value property-tax rate per $10,000 (tax), and lower status of the population (percent) (lstat).
A positive correlation between average number of rooms per dwelling (rm) with median value of owner-occupied homes in $1000s (medv).
A negative correlation between: lower status of the population (percent) (lstat) and median value of owner-occupied homes in $1000s (medv).
Moreover, three variables are negatively correlated with the weighted mean of distances to five Boston employment centres (dis). Those are proportion of owner-occupied units built prior to 1940 (age), nitrogen oxides concentration (parts per 10 million) (nox), and proportion of non-retail business acres per town (indus).
Summary of the variables
summary(Boston) ## crim zn indus chas
## Min. : 0.00632 Min. : 0.00 Min. : 0.46 Min. :0.00000
## 1st Qu.: 0.08205 1st Qu.: 0.00 1st Qu.: 5.19 1st Qu.:0.00000
## Median : 0.25651 Median : 0.00 Median : 9.69 Median :0.00000
## Mean : 3.61352 Mean : 11.36 Mean :11.14 Mean :0.06917
## 3rd Qu.: 3.67708 3rd Qu.: 12.50 3rd Qu.:18.10 3rd Qu.:0.00000
## Max. :88.97620 Max. :100.00 Max. :27.74 Max. :1.00000
## nox rm age dis
## Min. :0.3850 Min. :3.561 Min. : 2.90 Min. : 1.130
## 1st Qu.:0.4490 1st Qu.:5.886 1st Qu.: 45.02 1st Qu.: 2.100
## Median :0.5380 Median :6.208 Median : 77.50 Median : 3.207
## Mean :0.5547 Mean :6.285 Mean : 68.57 Mean : 3.795
## 3rd Qu.:0.6240 3rd Qu.:6.623 3rd Qu.: 94.08 3rd Qu.: 5.188
## Max. :0.8710 Max. :8.780 Max. :100.00 Max. :12.127
## rad tax ptratio black
## Min. : 1.000 Min. :187.0 Min. :12.60 Min. : 0.32
## 1st Qu.: 4.000 1st Qu.:279.0 1st Qu.:17.40 1st Qu.:375.38
## Median : 5.000 Median :330.0 Median :19.05 Median :391.44
## Mean : 9.549 Mean :408.2 Mean :18.46 Mean :356.67
## 3rd Qu.:24.000 3rd Qu.:666.0 3rd Qu.:20.20 3rd Qu.:396.23
## Max. :24.000 Max. :711.0 Max. :22.00 Max. :396.90
## lstat medv
## Min. : 1.73 Min. : 5.00
## 1st Qu.: 6.95 1st Qu.:17.02
## Median :11.36 Median :21.20
## Mean :12.65 Mean :22.53
## 3rd Qu.:16.95 3rd Qu.:25.00
## Max. :37.97 Max. :50.00Interpretations: On average,
- The per capital crime rate by the town is about 3.61.
- The proportion of residential land zoned for lots over 25,000 square feet is about 11.36.
- The proportion of non-retail business acres per town is about 11.14.
- The nitrogen oxides concentration (parts per 10 million) is about 0.55.
- The average number of rooms per dwelling is about 6.
- The proportion of owner-occupied units built prior to 1940 is about 68.57.
- The full-value property-tax rate per $10,000 is about 408.2.
- The pupil-teacher ratio by town is about 18.46.
- The median value of owner-occupied homes in $1000s is about 22.53.
- The chas Charles River dummy variable (= 1 if tract bounds river; 0 otherwise). Its mean is about 0.069.
Standardize the dataset
# center and standardize variables
boston_scaled <- scale(Boston)
# change the object to data frame
boston_scaled <- as.data.frame(boston_scaled)
# summaries of the scaled variables
summary(boston_scaled)## crim zn indus chas
## Min. :-0.419367 Min. :-0.48724 Min. :-1.5563 Min. :-0.2723
## 1st Qu.:-0.410563 1st Qu.:-0.48724 1st Qu.:-0.8668 1st Qu.:-0.2723
## Median :-0.390280 Median :-0.48724 Median :-0.2109 Median :-0.2723
## Mean : 0.000000 Mean : 0.00000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.007389 3rd Qu.: 0.04872 3rd Qu.: 1.0150 3rd Qu.:-0.2723
## Max. : 9.924110 Max. : 3.80047 Max. : 2.4202 Max. : 3.6648
## nox rm age dis
## Min. :-1.4644 Min. :-3.8764 Min. :-2.3331 Min. :-1.2658
## 1st Qu.:-0.9121 1st Qu.:-0.5681 1st Qu.:-0.8366 1st Qu.:-0.8049
## Median :-0.1441 Median :-0.1084 Median : 0.3171 Median :-0.2790
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.5981 3rd Qu.: 0.4823 3rd Qu.: 0.9059 3rd Qu.: 0.6617
## Max. : 2.7296 Max. : 3.5515 Max. : 1.1164 Max. : 3.9566
## rad tax ptratio black
## Min. :-0.9819 Min. :-1.3127 Min. :-2.7047 Min. :-3.9033
## 1st Qu.:-0.6373 1st Qu.:-0.7668 1st Qu.:-0.4876 1st Qu.: 0.2049
## Median :-0.5225 Median :-0.4642 Median : 0.2746 Median : 0.3808
## Mean : 0.0000 Mean : 0.0000 Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 1.6596 3rd Qu.: 1.5294 3rd Qu.: 0.8058 3rd Qu.: 0.4332
## Max. : 1.6596 Max. : 1.7964 Max. : 1.6372 Max. : 0.4406
## lstat medv
## Min. :-1.5296 Min. :-1.9063
## 1st Qu.:-0.7986 1st Qu.:-0.5989
## Median :-0.1811 Median :-0.1449
## Mean : 0.0000 Mean : 0.0000
## 3rd Qu.: 0.6024 3rd Qu.: 0.2683
## Max. : 3.5453 Max. : 2.9865How did the variables change?: The variables have been rescaled to have a mean of zero and a standard deviation of one. For a standardized variable, each case’s value on the standardized variable indicates it’s difference from the mean of the original variable in number of standard deviations (of the original variable).
# Create a categorical variable of the crime rate
# create a quantile vector of crim and print it
bins <- quantile(boston_scaled$crim)
# create a categorical variable 'crime'
crime <- cut(boston_scaled$crim, breaks = bins, include.lowest = TRUE, label = c("low", "med_low", "med_high", "high"))
# look at the table of the new factor crime
table(crime)## crime
## low med_low med_high high
## 127 126 126 127# Drop the old crime rate variable from the dataset
boston_scaled <- dplyr::select(boston_scaled, -crim)
# add the new categorical value to scaled data
boston_scaled <- data.frame(boston_scaled, crime)
# Divide the dataset to train and test sets
# number of rows in the Boston dataset
n <- nrow(boston_scaled)
# choose randomly 80% of the rows
ind <- sample(n, size = n * 0.8)
# create train set
train <- boston_scaled[ind,]
# create test set
test <- boston_scaled[-ind,]
# save the correct classes from test data
correct_classes <- test$crime
# remove the crime variable from test data
test <- dplyr::select(test, -crime)Fit the linear discriminant analysis on the train set
# linear discriminant analysis
lda.fit <- lda(crime ~ ., data = train)
# the function for lda biplot arrows
lda.arrows <- function(x, myscale = 1, arrow_heads = 0.1, color = "red", tex = 0.75, choices = c(1,2)){
heads <- coef(x)
arrows(x0 = 0, y0 = 0,
x1 = myscale * heads[,choices[1]],
y1 = myscale * heads[,choices[2]], col=color, length = arrow_heads)
text(myscale * heads[,choices], labels = row.names(heads),
cex = tex, col=color, pos=3)
}
# target classes as numeric
classes <- as.numeric(train$crime)
# plot the lda results
plot(lda.fit, dimen = 2, col = classes, pch = classes)
lda.arrows(lda.fit, myscale = 1.3)
Prediction of the classes with the LDA model on the test data
# Save the crime categories from the test set and then remove the categorical crime variable from the test dataset. DONE -- See above steps.
# predict classes with test data
lda.pred <- predict(lda.fit, newdata = test)
# cross tabulate the results
tbl_lda <- table(correct = correct_classes, predicted = lda.pred$class)
tbl_lda; rowSums(tbl_lda)## predicted
## correct low med_low med_high high
## low 14 6 3 0
## med_low 9 16 3 0
## med_high 2 13 13 0
## high 0 0 0 23## low med_low med_high high
## 23 28 28 23Comments: From the table, we see that the LDA predicts correctly:
- 16 out of 28 (i.e., 57.1%) low.
- 21 out of 26 (i.e., 80.8%) med_low.
- 18 out of 22 (i.e., 81.8%) med_high.
- All 26 (i.e., 100%) high.
Reload the dataset, standardize and run k-means algorithm
data("Boston") # Reload
Re_data <- scale(Boston) # standardize it
# distances between the observations
dist_eu <- dist(Re_data)
# k-means clustering with 3 clusters
km <- kmeans(Re_data, centers = 3)
# investigate what is the optimal number of clusters
set.seed(123) # set the seed
# determine the number of clusters
k_max <- 10
# calculate the total within sum of squares
twcss <- sapply(1:k_max, function(k){kmeans(Re_data, k)$tot.withinss})
# visualize the results
library(ggplot2)
qplot(x = 1:k_max, y = twcss, geom = 'line')
Comment: The “scree plot” above helps to identify the appropriate number of clusters. The “elbow shape” suggests that to clusters (k = 2) is the potential candidate, since the total WCSS drops radically.
# run the algorithm again
# k-means clustering
km_new <- kmeans(Re_data, centers = 2)
# plot the Re_scale Boston dataset with 2 clusters
pairs(Re_data, col = km_new$cluster)
table(km_new$cluster) ##
## 1 2
## 329 177Comment: With k = 2, clusters consist of 329 observations out of 506 in cluster 1, and cluster 2, 177 observations. The clusters are also separated by colour within the predictors. Some variables present a clear cut of observations, other it is a quite mix.
Bonus
model_predictors <- dplyr::select(train, -crime)
# check the dimensions
dim(model_predictors)## [1] 404 13dim(lda.fit$scaling)## [1] 13 3# matrix multiplication
matrix_product <- as.matrix(model_predictors) %*% lda.fit$scaling
matrix_product <- as.data.frame(matrix_product)
library(plotly) # access plotly library
# 3D plot of the columns of the matrix
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', surfacecolor = train$crime)# another 3D plot with color defined by the clusters of the k-means
plot_ly(x = matrix_product$LD1, y = matrix_product$LD2, z = matrix_product$LD3, type= 'scatter3d', mode='markers', surfacecolor = km_new$cluster)How do the plots differ? Are there any similarities?: The two plots look relatively similar with a clear cut between clusters, and the number of observation within each group seems to be the same.